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Let $z=e^{\frac{2\pi i}{5}}$, then $1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=?$

I am kind of confused since by drawing a graph, $1+z+z^2+z^3+z^4$ should be zero, but using computational softwares the result is different, and hence I do not know how to solve this problem. Thanks for helping!

Amber Xue
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  • I agree that $1+z+z^2+z^3+z^4=0$ and suspect that you've made some little mistake with your software computation. – Mark McClure Aug 27 '14 at 15:01
  • seems that I cannot upload a photo of the computation output by R...the sum equalling 0 seems more reasonable for me, too – Amber Xue Aug 27 '14 at 15:03
  • I don't know R but, if you're working symbolically, try (1) simplifying the result and (2) evaluate it numerically. – Mark McClure Aug 27 '14 at 15:04
  • yeah...simplified finally to $5e^{\frac{18\pi i}{5}}$! – Amber Xue Aug 27 '14 at 15:07
  • Since $1,z,z^2,z^3,z^4$ are the roots of $z^5-1=0$, they have to sum to $0$ since the next-to-leading coefficient of that polynomial is $0$.

    Even better, the sum $S=1+z+ \ldots z^5$ satisfies $zS=z$. Since $z$ is not $0$, $S=0$.

     – Dylan Yott Aug 27 '14 at 15:21

2 Answers2

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According to wolframalpha you are right about $1+z+z^2+z^3+z^4 = 0$

For the sum :

$1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9$

$(1+z+z^2+z^3+z^4)+4z^4+4z^5+4z^6+4z^7+4z^8+5z^9$

$0+4z^4(1+z+z^2+z^3+z^4)+5z^5*z^4$

$4z^4(0)+5*1*z^4$

$5*z^4$

GuiguiDt
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We have $$1+z+z^2+z^3+z^4+5z^4+4z^5+4z^6+4z^7+4z^8+4z^8+5z^9=$$

$$=(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9.$$

Using the fact that the $\left(e^{2\pi i/5}\right)^5=1$, we see that $z=e^{2\pi i/5}$ is a root of $z^5-1=(z-1)(1+z+z^2+z^3+z^4)$, in particular, it must be a root of $1+z+z^2+z^3+z^4$, so that we are left with $$5z^9=5e^{(2\pi i/5)\cdot 9}=5e^{18\pi i/5}=5e^{8\pi i/5},$$where we have used the fact that $e^{2\pi i}=1$.

Clayton
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